Arecent article reported that a job awaits only one in three new college graduates. the major reasons given were an overabundance of college graduates and a weak economy. a survey of 200 recent graduates from your school revealed that 80 students had jobs. at the 0.01 significance level, can we conclude that a larger proportion of students at your school have jobs?

If we compare the p value obtained and the significance level given we have so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion of students that had job at the school mentioned is not significantly higher then 0.33 .  

Step-by-step explanation:

1) Data given and notation  

n=200 represent the random sample taken

X=80 represent the number of students that had jobs

estimated proportion of students that had jobs

is the value that we want to test

represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion of students that have jobs at the school mentioned is higher than the reported value at the article:  

Null hypothesis:  

Alternative hypothesis:  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

(1)  

The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value .

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided . The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

If we compare the p value obtained and the significance level given we have so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion of students that had job at the school mentioned is not significantly higher then 0.33 .  

the answer is 510

step-by-step explanation:

15 • 13 = 195

15 • 12 = 180

15 • 5 = 75

12 • 5 = 60

195 + 180 = 375

375 + 75 = 450

450 + 60 = 510

lockers 1, 4, 9, 16, 25, 40, and 49 are open

i ran through it by the other answer is wrong